This is not a partial

fraction. Maybe a trig substitution.

[tex]\int\sqrt{x^{2}+2x}dx=\int\sqrt{x(x+2)}dx[/tex]

This one can be a little tricky though.

Let [tex]x=u-1, \;\ u+1=x+2, \;\ du=dx[/tex]

Make the subs and we have [tex]\int\sqrt{(u-1)(u+1)}du=\int\sqrt{u^{2}-1}du[/tex]

Now, it is a form in which we can use trig sub easily enough.

[tex]\int\sqrt{u^{2}-1}du[/tex]

Let [tex]u=sec(t), \;\ du=sec(t)tan(t)dt[/tex]

Make the subs:

[tex]\int\sqrt{sec^{2}(t)-1}sec(t)tan(t)dt[/tex]

But [tex]sec^{2}(t)-1=tan^{2}(t)[/tex]

[tex]\int\sqrt{tan^{2}(t)}sec(t)tan(t)dt[/tex]

[tex]\int tan^{2}(t)sec(t)dt[/tex]

[tex]\int sec(t)(sec^{2}(t)-1)dt[/tex]

[tex]\int sec^{3}(t)dt-\int sec(t)dt[/tex]

Now, I will let you finish. OKey-doke?. Use the reduction formula if need be on the sec and you have it. That formula is in any calc book

See?. No partial fractions. When you get that integrated, you will need to resub to get it back in terms of x.

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